Question

A moving coil galvanometer has 100 turns and each turn has an area of $2.0 \, \text{cm}^2$. The magnetic field produced by the magnet is $0.01 \, \text{T}$ and the deflection in the coil is $0.05$ radians when a current of $10 \, \text{mA}$ is passed through it. The torsional constant of the suspension wire is $x \times 10^{-5} \, \text{N} \cdot \text{m/rad}$. The value of $x$ is:

Answer 1

Correct Answer: 4

Given: - Number of turns: \(N = 100\) - Area of each turn: \(A = 2.0 \, \text{cm}^2 = 2.0 \times 10^{-4} \, \text{m}^2\) - Magnetic field: \(B = 0.01 \, \text{T}\) - Deflection: \(\theta = 0.05 \, \text{radian}\) - Current: \(I = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A}\)

Step 1: Calculating the Torque Acting on the Coil

The torque (\(\tau\)) acting on a moving coil galvanometer is given by:

\[ \tau = N \times B \times I \times A \]

Substituting the given values:

\[ \tau = 100 \times 0.01 \, \text{T} \times 10 \times 10^{-3} \, \text{A} \times 2.0 \times 10^{-4} \, \text{m}^2 \] \[ \tau = 100 \times 0.01 \times 0.01 \times 2.0 \times 10^{-4} \, \text{N} \times \text{m} \] \[ \tau = 2 \times 10^{-5} \, \text{N} \times \text{m} \]

Step 2: Relating Torque to the Torsional Constant

The torque is also related to the torsional constant (\(k\)) and deflection (\(\theta\)) by:

\[ \tau = k \times \theta \]

Rearranging to find \(k\):

\[ k = \frac{\tau}{\theta} \]

Substituting the given values:

\[ k = \frac{2 \times 10^{-5}}{0.05} \, \text{N} \times \text{m/rad} \] \[ k = 4 \times 10^{-4} \, \text{N} \times \text{m/rad} \]

Step 3: Finding the Value of \(x\)

Given that \(k = x \times 10^{-5} \, \text{N} \times \text{m/rad}\):

\[ 4 \times 10^{-4} = x \times 10^{-5} \]

Solving for \(x\):

\[ x = 4 \]

Conclusion: The value of \(x\) is 4.